Answer:
Step-by-step explanation:
Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?
Let p(x)=kpx+dp and q(x)=kqx+dq than
f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7
p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7
(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)
So you want:
−2kpk2q=0
and
kpkq=−1
and
2kpd2p−3kpdq+7=0
Now I amfraid this doesn’t work as −2kpk2q=0 that either kp or kq is zero but than their product can’t be anything but 0 not −1 .
Answer: there are no such linear functions.
Answer:
x=112
Step-by-step explanation:
x/2-18=38
first add +18
x/2=56
now multiply by 2
x=112
Answer:
x = 4
Step-by-step explanation:
√x + 1 = 3
√x = 3 - 1
√x = 2
(√x)² = (2)²
x = 4
Answer:
third one
Step-by-step explanation:
Answer:
79.5 + 5.5x = Y
Step-by-step explanation:
Sumo wrestler gained 5.5 kg per month
After 11 month, he weighed 140 kg.
Let x be his current weight.
Then x + 11(5.5) = 140
x = 140 - 60.5
x = 79.5
If Y is the weight of the wrestler after t months, then the linear equation would be:
79.5 + 5.5t = Y
